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Solution 2.1:3e

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Both terms contain x, which can therefore be taken out as a factor (as can 2),

18x−2x3=2x9−2xx2=2x(9−x2).

The remaining second-degree factor 9−x2 can then be factorized using the conjugate rule

2x(9−x2)=2x(32−x2)=2x(3+x)(3−x)

which can also be written as −2x(x+3)(x−3)

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